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3 years ago

Consider a horse pulling a buggy. Is the

Physics

1 answer:

Degger [83]3 years ago

8 0

Yes, it's true.

But 2nd Newton Law always come to play when the horse is to move forward because obviously the forces interact antagonistically and mass has to be accounted for.

That's what I think. Hope it's right, all the best.

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A 68.2 kg base runner begins his slide into second base while moving at a speed of 4.33 m/s. He slides so that his speed is zero

qwelly [4]

Answer:

$147.653J = W_f$

Explanation:

We know that:

$E_i - E_f = W_f$

where $E_i$ is the initial energy, $E_f$ is the final energy and $W_f$ is the energy lost due to friction.

so:

$\frac{1}{2}MV^2-0 = W_f$

Where M is the mass and V is the velocity of the runner, so:

$\frac{1}{2}(68.2kg)(4.33m/s)^2 = W_f$

$147.653J = W_f$

7 0

4 years ago

Differentiate between loadstone and bar magnet

MakcuM [25]

As nouns the difference between magnet and lodestone

is that magnet is a piece of material that attracts some metals by magnetism while lodestone is a naturally occurring magnet.

7 0

2 years ago

Aranza is 17 years old. What is her Maximum Heart Rate? I will give brainliest

tensa zangetsu [6.8K]

Answer:

203 beats per minute

Explanation:

4 0

3 years ago

Read 2 more answers

A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the t

cricket20 [7]

Answer:

Part a)

here friction force will accelerate the box in forward direction

Part b)

a = 3.14 m/s/s

Explanation:

Part a)

When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction

Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck

So here friction force will accelerate the box in forward direction

Part b)

The maximum value of friction force on the box is known as limiting friction

it is given by the formula

$F = \mu mg$

so we have

$F = ma = \mu mg$

now the acceleration is given as

$a = \mu g$

$a = (0.32)(9.8) = 3.14 m/s^2$

5 0

3 years ago

A ball starts at rest and rolls down a ramp, developing a speed of 45 m/s in 7 seconds. Calculate the ball's acceleration.

Alexxx [7]

Answer:

6.429 m/s^2.

Explanation:

Using equations of motion,

i. vf = vi + at

ii. vf^2 = vi^2 + 2a*S

iii. S = vi*t + 1/2 * (a*t^2)

Where,

vf = final velocity of the motion

vi = initial velocity of the motion

S = distance travelled

t = time taken to complete the motion

a = acceleration due to gravity

Given:

vi = 0m/s

vf = 45 m/s

t = 7 s

a = ?

Using the i. equation of motion,

vf = vi + at

45 = 0 + a*7

a = 45/7

= 6.429 m/s^2

6 0

3 years ago